The racing cars of today are equipped with so many sophisticated equipment. Introduction of a new
visual transducer which is interfaced with the on-board computer can tell you on the fly how many
cars are ahead of you while how many are trailing. There are N cars on a racing track. Each has an
on-board computer with the new feature. During the race, every single car’s computer keeps displaying
two integers, a (The number of cars in front of him) & b (The number of cars behind him) for a
particular moment. It is possible that at some time, some of the cars are racing side by side i.e. they
are exactly at the same location. A car will not consider any other car at the same location to be a
leading or trailing car.
Now, it is suspected that some of the new transducers are not working properly inside such high
speed vehicles. The report with all computers’ data generated at a particular timestamp is reported to
you. You are to determine the minimum number of cars that have faulty data.
Input
Each test case begins with an integer N (1 ≤ N ≤ 1000), the number of cars on a track. The next N
lines each has two integers — a & b (0 ≤ a, b ≤ 1500) for a particular car.
The last test case is followed by a line with a single ‘0’ indicating the end of input.
Output
For each test case, print a line in the format, ‘Case X: Y ’, where X is the case number & Y is the
minimum number of cars that must have faulty data according to the report.
Sample Input
4
2 2
0 0
0 2
3 1
1
1 1
0
Sample Output
Case 1: 3
Case 2: 1

题目大概：

给出n辆车，每辆车给出在它前面的车的数量和在它后面的车的数量，然后求最少有多少车给出的位置信息是错误的。 

思路：

可以从反面考虑，那就是求最多的合法的车的数量。

因为给出了前面a 辆车，后面 b 辆车，所以它一定在一个区间范围内，【a+1，n-b】，并且是在任何一个位置都应该没有问题，如果两辆车合法的话，它们的区间一定不会相交，因为会有冲突，即如果两练车有相交的部分，我们在求解合法车的时候，选了1号，就不能选2号。所以合法的区间一定相离，那么就用区间dp，枚举每个区间就行了。

但有一个地方要注意，就是一个区间【j，i】，最多的车的数量是i-j+1，因为在这个区间最多放这些车，就是最多有并排的这些车。

代码：

#include <bits/stdc++.h>

using namespace std;
const int maxn=1050;
int a[maxn][maxn];
int dp[maxn];
int main()
{
    int n;
    int ans=1;
    while(scanf("%d",&n)&&n)
    {
        int u,v;
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&u,&v);
            a[u+1][n-v]++;
        }
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<i;j++)
            {
                int max_1=i-j;
                dp[i]=max(dp[i],min(dp[j]+a[j+1][i],dp[j]+max_1));
            }
        }
        printf("Case %d: %d\n",ans++,n-dp[n]);
    }
    return 0;
}